Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define maxn 510
#define ll long long
#define db double
#define vi vector<int>
#define pb push_back
#define mod 998244353
#define fi first
#define pi pair<int, int>
#define se second
#define mp make_pair
ll ksm(ll a, ll b) {
   if (!b) return 1;
   ll ns = ksm(a, b >> 1);
   ns = ns * ns % mod;
   if (b & 1) ns = ns * a % mod;
   return ns;
}
// head
ll a[maxn][maxn];
const int S = 20050;
bool bpr[S];
map<int, int> r[2];
ll gcd(ll a, ll b) {
	if (!b) return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}	
int main() {
	vi pr;
	for (int i = 0; i < S; i++)
		bpr[i] = 1;
	for (int i = 2; i < S; i++) {
		if (!bpr[i]) continue;
		pr.pb(i);
		for (int j = 2; j * i < S; j++)
			bpr[i * j] = 0;
	}
	int n;
	cin >> n;
	if (n == 2) {
		cout << 4 << ' ' << 7 << endl << 23 << ' ' << 10 << endl;
		return 0;
	}
	int cnt = 0;
	for (int i = 1; i <= 2 * n; i += 2)
		r[0][i] = pr[cnt++];
	for (int i = -n - 1; i <= n + 1; i++)
		if (i % 2)
			r[1][i] = pr[cnt++];
 	for (int i = 1; i <= n; i++) {
 		for (int j = 1; j <= n; j++) {
 			if ((i + j) % 2 == 0) {
	 			ll ns = 1ll * max(1, r[1][i - j + 1]) * max(1, r[1][i - j - 1]) * max(1, r[0][i + j + 1]) * max(1, r[0][i + j - 1]);
	 			a[i][j] = ns + 1;
			}
			else a[i][j] = r[0][i + j] * r[1][i - j];
			printf("%lld ", a[i][j]);
		}
		printf("\n");
	}
	return 0;
}